In a survey 6000 women 3431 say they change their nail polis

In a survey 6000 women. 3431 say they change their nail polish once week. Construct a 90% confidence interval for the population proportion of women who can change their nail polish once weekly. 90% confidence interval for the population proportion is?

Solution

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=3431
Sample Size(n)=6000
Sample proportion = x/n =0.5718
Confidence Interval = [ 0.5718 ±Z a/2 ( Sqrt ( 0.5718*0.4282) /6000)]
= [ 0.5718 - 2.58* Sqrt(0) , 0.5718 + 2.58* Sqrt(0) ]
= [ 0.5553,0.5883]